If the curves x^{2}-6x+y^{2}+8=0 and x^{2}-8y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given equations of the circles are $S_{1}: x^{2}+y^{2}-6x+8=0$ and $S_{2}: x^{2}+y^{2}-8y+16-k=0$.For $S_{1}$,the center $C_{1} = (3, 0)$ and

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) The given equations of the circles are $S_{1}: x^{2}+y^{2}-6x+8=0$ and $S_{2}: x^{2}+y^{2}-8y+16-k=0$.For $S_{1}$,the center $C_{1} = (3, 0)$ and radius $r_{1} = \sqrt{3^{2}+0^{2}-8} = \sqrt{1} = 1$.For $S_{2}$,the center $C_{2} = (0, 4)$ and radius $r_{2} = \sqrt{0^{2}+4^{2}-(16-k)} = \sqrt{k}$.The distance between the centers is $d = \sqrt{(3-0)^{2}+(0-4)^{2}} = \sqrt{9+16} = 5$.Since the circles touch each other,the distance between centers must be equal to the sum or difference of the radii: $d = |r_{1} \pm r_{2}|$.Case $1$: $d = r_{1}+r_{2}$ $\Rightarrow 5 = 1+\sqrt{k}$ $\Rightarrow \sqrt{k} = 4$ $\Rightarrow k = 16$.Case $2$: $d = |r_{1}-r_{2}| \Rightarrow 5 = |1-\sqrt{k}|$.This implies $1-\sqrt{k} = 5$ or $1-\sqrt{k} = -5$.$1-\sqrt{k} = 5 \Rightarrow \sqrt{k} = -4$ (not possible as $k>0$).$1-\sqrt{k} = -5$ $\Rightarrow \sqrt{k} = 6$ $\Rightarrow k = 36$.Comparing the values,the largest value of $k$ is $36$.

If the curves $x^{2}-6x+y^{2}+8=0$ and $x^{2}-8y+y^{2}+16-k=0$ $(k>0)$ touch each other at a point,then the largest value of $k$ is

Similar Questions

The equations of three circles are $x^2 + y^2 - 12x - 16y + 64 = 0$,$3x^2 + 3y^2 - 36x + 81 = 0$,and $x^2 + y^2 - 16x + 81 = 0$. The coordinates of the point from which the lengths of the tangents drawn to each of the three circles are equal is:

Suppose that the circle $x^2+y^2+2gx+2fy+c=0$ has its centre on $2x+3y-7=0$ and cuts the circles $x^2+y^2-4x-6y+11=0$ and $x^2+y^2-10x-4y+21=0$ orthogonally. Then $5g-10f+3c=$

If $(a, b)$ is the common point for the circles $x^2+y^2-4x+4y-1=0$ and $x^2+y^2+2x-4y+1=0$,then $a^2+b^2=$

If the origin lies on a diameter of the circle $x^2+y^2-4x-2y-4=0$,then the equation of the circle passing through the end points of that diameter and the point $(1,2)$ is

Find the value of $m+n$,if the circumference of the circle $x^2+y^2+8x+8y-m=0$ is bisected by the circle $x^2+y^2-2x+4y+n=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module